Question

Compute the definite integrals. Use a graphing utility to confirm your answers. $$ \int_{0}^{1} e^{\sqrt{x}} d x(\operatorname{let} u=\sqrt{x}) $$

Short answer

The definite integral evaluates to 2.

Step-by-step solution

Substitution

First, we perform a substitution. Let \( u = \sqrt{x} \), then \( x = u^2 \). When \( x = 0 \), \( u = 0 \), and when \( x = 1 \), \( u = 1 \). Now, differentiate \( x = u^2 \) to find \( dx \); we have \( dx = 2u \, du \).

Change Limits of Integration

Rewrite the integral with the new variable \( u \): \[ \int_{0}^{1} e^{\sqrt{x}} \, dx = \int_{0}^{1} e^u \, 2u \, du \] This utilizes our substitution \( u = \sqrt{x} \) and \( dx = 2u \, du \).

Integral of the New Expression

Now we need to integrate the expression \( 2u e^u \), which requires integration by parts. Let \( v = e^u \) and \( dv = e^u \, du \), choose \( w = u \) and \( dw = du \). Apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, dw \] Substitute the values: \[ \int u \, e^u \, du = u e^u - \int e^u \, du \] Therefore: \[ \int e^u (u) \, du = e^u (u - 1) \]

Evaluate Definite Integral

Now evaluate the definite integral from 0 to 1: \[ 2 \int_{0}^{1} e^u u \, du = 2 [e^u (u - 1)]_{0}^{1} \] Evaluate: \[ = 2 \left[ (1 e^1 - e^1) - (0 - e^0) \right] = 2 \left[ e - e + 1 \right] = 2 \times 1 = 2\]

Confirm with Graphing Utility

Use a graphing utility, such as Desmos or a graphing calculator, to confirm the integral calculated. Enter the integral \( \int_{0}^{1} e^{\sqrt{x}} \, dx \) and verify the findings against the graphs or numerical solutions given by the utility.

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. Unlike indefinite integrals, which represent families of functions, definite integrals give us a specific numerical value. This value represents the net area between the graph of the function and the x-axis over a given interval.
A definite integral can be written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. The process involves evaluating the antiderivative of the function at these points and finding the difference.
Definite integrals have numerous applications, from calculating areas to determining quantities like distance, mass, and more. When working with definite integrals, it's common to use various techniques like substitution or integration by parts, especially when we're dealing with more complicated functions.

Integration by Parts

Integration by parts is a method used to evaluate integrals where the standard antiderivative approach doesn't work directly. This technique is grounded in the product rule for differentiation and is particularly useful for integrals of products of functions.
The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). To apply this, you choose parts of your function to represent \( u \) and \( dv \). Then, differentiate \( u \) to find \( du \) and integrate \( dv \) to get \( v \).
This method transforms a complicated integral into hopefully simpler integrals, making it easier to solve. It's important to choose \( u \) and \( dv \) in a manner that simplifies the problem, a process that sometimes involves some trial and error.
In our solution, we applied integration by parts to the expression \( \int u e^u \, du \), letting \( u \) simplify the integration process by transforming it to a more manageable form.

Substitution Method

The substitution method is another powerful tool in calculus for simplifying integrals, particularly when you have a composite function. It works by changing variables to simplify the integral into an easier form that can be directly integrated.
This involves choosing a substitution \( u = g(x) \), which simplifies the integration. Calculate \( du = g'(x) dx \) and substitute back into the integral. Additionally, the limits of integration also need to be adjusted according to the new variable.
In the context of our problem, we used the substitution \( u = \sqrt{x} \) to simplify the integrand \( e^{\sqrt{x}} \). This transforms the integral into an equivalent form that's hopefully easier to evaluate.
Substitution is a bit like a change of perspective; by reframing the problem, it often becomes more approachable, allowing you to find the antiderivative more effectively. This method is particularly useful when dealing with nested or intricate functions involving compositions or complicated expressions.